## Chapter 6 Squares and Square Roots Exercise 6.2

**Question 1: Find the square of the following numbers.**

i. 32

ii. 35

iii. 86

iv. 93

v. 71

vi. 46

Answer:

i. 32

ii. 35

iii. 86

iv. 93

v. 71

vi. 46

Answer:

i.

32 = 30 + 2

32² = 32 × 32

32² = (30 + 2)² = 30(30 + 2) + 2(30 + 2)

= 30² + 30 × 2 + 2 × 30 + 2²

= 900 + 60 + 60 + 4

= 1024

ii.

35 = 30 + 5

35² = 35 × 35

35² = (30 + 5)² = 30(30 + 5) + 5(30 + 5)

= 30² + 30 × 5 + 5 × 30 + 5²

= 900 + 150 + 150 + 25

= 1225

iii.

86 = 80 + 6

86² = 86 × 86

86² = (80 + 6)² = 80(80 + 6) + 6(80 + 6)

= 80² + 80 × 6 + 6 × 80 + 6²

= 6400 + 480 + 480 + 36

= 7396

iv.

93 = 90 + 3

93² = 93 × 93

93² = (90 + 3)² = 90(90 + 3) + 3(90 + 3)

= 90² + 90 × 3 + 3 × 90 + 3²

= 8100 + 270 + 270 + 9

= 8649

v.

71 = 70 + 1

71² = 71 × 71

71² = (70 + 1)² = 70(70 + 1) + 1(70 + 1)

= 70² + 70 × 1 + 1 × 70 + 1²

= 4900 + 70 + 70 + 1

= 5041

vi.

46 = 40 + 6

46² = 46 × 46

46² = (40 + 6)² = 40(40 + 6) + 6(40 + 6)

= 40² + 40 × 6 + 6 × 40 + 6²

= 1600 + 240 + 240 + 36

= 2116

**Question 2: Write a Pythagorean triplet whose one member is**

i. 6

ii. 14

iii. 16

iv. 18

Answer:

i. 6

ii. 14

iii. 16

iv. 18

Answer:

i.

2m = 6

m = 6/2

m = 3

So, 2m = 2 × 3 = 6

m² - 1 = 3² - 1 = 9 - 1 = 8

m² + 1 = 3² + 1 = 9 + 1 = 10

Therefore, the Pythagoras triplet is 6, 8, 10.

ii.

2m = 14

m = 14/2

m = 7

So, 2m = 2 × 7 = 14

m² - 1 = 7² - 1 = 49 - 1 = 48

m² + 1 = 7² + 1 = 49 + 1 = 50

Therefore, the Pythagoras triplet is 14, 48, 50.

iii.

2m = 16

m = 16/2

m = 8

So, 2m = 2 × 8 = 16

m² - 1 = 8² - 1 = 64 - 1 = 63

m² + 1 = 8² + 1 = 64 + 1 = 65

Therefore, the Pythagoras triplet is 16, 63, 65.

iv.

2m = 18

m = 18/2

m = 9

So, 2m = 2 × 9 = 18

m² - 1 = 9² - 1 = 81 - 1 = 80

m² + 1 = 9² + 1 = 81 + 1 = 82

Therefore, the Pythagoras triplet is 18, 80, 82.

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